Report Shearwall
-
Upload
roy-ascencio-sanabria -
Category
Documents
-
view
227 -
download
1
Transcript of Report Shearwall
UNIVERSIDAD NACIONAL DE INGENIERÍA
FACULTAD DE INGENIERIA CIVIL
CENTRO PERUANO JAPONES DE INVESTIGACIONES
SISMICAS Y MITIGACION DE DESASTRES
Investigations of Shear Walls Matthias Schmalzbauer
May 2011
General
In structural engineering, a shear wall is a wall composed of braced panels to counter the effects of lateral load acting on a structure. Wind and earthquake loads are the most common loads braced wall lines are designed to counteract.
A fast way analyzing a concrete wall for large eccentricities is the “Interaction Diagram”. The Diagram is a curve plot of points, where each point hast two ordinates. The first ordinate is the bending moment strength and the second one is the corresponding axial force. Both ordinates are linked with eccentricity. The shape of the curve can be defined by finding the ordinates of major points.
General calculation of P and M
c
dn
εc
εc εsu+( ) cεc
εc εsu+( ) dn⋅
fsi
fy
εsuεsi⋅ 4200
kgf
cm2
≤ 4200−kgf
cm2
≥
εsi
c di−( )c
εc⋅
β 0.85f¨c 200−( )
700.05⋅−
Fsi fsi Asi⋅
Cc 0.85 f´c⋅ B⋅ a⋅
a β c⋅P Cc
1
n
i
Fsi∑=
+
M CcH
2
a
2−
⋅
1
n
i
FsiH
2di−
⋅
∑
=
+
Boundary conditions:
0.85
0.65
β
fc
kgf
cm2
280 560
Diagram for finding β
Section
Deformation at the “Balanced condition”
Forces at the “Balanced Condition”
Finding c at “Balanced Condition”:
Note: Balanced Condition:
Tension failure = Compression failure
Deformation:
Forces:
Resulting Force:
(Balance of Forces)
q
(Moments around q)
Main points “Interaction Diagram”
Point 1: Axial compression at zero moment. Point 2: Maximum axial compression load permitted by code at zero eccentricity. Point 3: Maximum moment strength at the maximum axial compression load permitted by code. Point 4: Compression and moment strength at balanced conditions. Point 5: Moment strength at zero axial force.
-100
0
100
200
300
400
500
600
0 20 40 60 80 100 120 140 160
Ax
ial
Co
mp
ress
ion
P [
t]
Bending Moment M [t*m]
1
2
3
4
5
comp. control
tension control
balanced
Strain distribution corresponding to points on the "Interaction Diagram"
Any combination of P and M outside the envelope will cause failure
Strain distribution corresponding to points on the "Interaction Diagram"
Any combination of P and M outside the envelope will cause failure
Concrete failure
Steel failure
Strain distribution corresponding to points on the "Interaction Diagram"
Any combination of P and M outside the envelope will cause failure
Calculation
Shear-Wall Type A: Rectangular shape
(Calculation - Excel)
Material Definition: (Entered Data)
Geometric Data: (Entered Data)
Illustrated shape Actual arrangement
(β changes with change of f`c)
(Yield strength)
(Compressive strength)
(Strain concrete)
(Yield strain steel)
ΣA si
H B⋅( )
Asiπ4
φ i
8
2.54⋅ cm⋅
2
⋅
cB
εc
εc εsu+( ) dn⋅
aB β cB⋅
fsi
fy
εsuεsi⋅
Fsi fsi Asi⋅
Msi FsiH
2d−
⋅
M CcH
2
a
2−
⋅
1
n
i
FsiH
2di−
⋅
∑
=
+ P Cc
1
n
i
Fsi∑=
+
Cc 0.85 f´c⋅ B⋅ aB⋅
εsi
cB di−( )cB
εc⋅
Reinforcement:
Calculation Balanced-Point: (Point 4) Note: Getting more points by changing cB
Interaction Diagram Shear-Wall Type A: (Rectangular shape)
-100
0
100
200
300
400
500
600
0 20 40 60 80 100 120 140 160
Ax
ial
Fo
rce
P [
t]
Bending Moment M [t*m]
Pmax 0.85 f'c⋅ B H⋅
1
n
i
Asi∑=
−
⋅
1
n
i
Asi∑=
fy⋅+
Further Results with changed c: (Axial Force “P” and Bending Moment “M”)
Calculation: (Point 1) - Axial compression at zero moment
(Point 1)
(Point 4)
Shear-Wall Type B: Wall with two columns
(Calculation - Excel)
Material Definition: (Entered Data)
Geometric Data: (Entered Data)
Illustrated shape Actual arrangement
(β changes with change of f`c)
(Yield strain steel)
(Strain concrete)
(Yield strength)
(Compressive strength)
ΣA si
H B⋅( )
Asiπ4
φ i
8
2.54⋅ cm⋅
2
⋅
Pmax 0.85 f'c⋅ B H⋅
1
n
i
Asi∑=
−
⋅
1
n
i
Asi∑=
fy⋅+
Reinforcement:
Calculation: (Point 1) - Axial compression at zero moment
cB
εc
εc εsu+( ) dn⋅
aB β cB⋅
fsi
fy
εsuεsi⋅
Fsi fsi Asi⋅
Msi Fsi
2 H2⋅ H1+( )2
d−
⋅
Cc 0.85 f´c⋅ B2⋅ a⋅
P Cc
1
n
i
Fsi∑=
+
εsi
cB di−( )cB
εc⋅
M Cc
H1 2 H2⋅+( )2
a
2−
⋅
1
n
i
Fsi
2 H2⋅ H1+( )2
⋅ di−
∑
=
+
Calculation Balanced-Point: (Point 4) Note: Getting more points by changing cB
Interaction Diagram Shear-Wall Type B: (Wall with two columns)
-500
0
500
1000
1500
2000
0 100 200 300 400 500 600 700
Ax
ial
Fo
rce
P [
t]
Bending Moment M [t*m]
Further Results with changed c: (Axial Force “P” and Bending Moment “M”)
(Point 1)
(Point 4)
Calculation
Shear-Wall Type A: Rectangular shape
(Mathcad)
Material definitions
E-Module Concrete - (Kent and Park)
Concrete under multiaxial compressive stress state exhibits significant nonlinearity, which can be successfully represented by nonlinear constitutive models and is characterized by parabolic stress-strain relationship. Elastic limit strain and strain at cracking are limited to 0.2%.
Entered data:
fc 0.85 280⋅kgf
cm2
:=
fc 238kgf
cm2
⋅=
f εc( ) fc 2εc
0.002⋅
εc0.002
2
−
⋅ εc 0.002≤if
fc 1 100 εc 0.002−( )⋅−[ ]⋅ εc 0.002>if
:=
0 1 103−
× 2 103−
× 3 103−
× 4 103−
×0
1 107
×
2 107
×
3 107
×
f εc( )
εc
Stress-strain graphic for concrete (Kent and Park)
(Maximum stress of concrete)
0.85: Standarts
E-Module Steel - (Bilinear diagram with parable – Park and Paulay)
Steel is isotropic and homogeneous material exhibiting stress-strain relationship. While the ultimate limit strain in tension and that of compression are taken as 1% and 0.35%, respectively, elastic strain in steel in tension and compression are considered the same.
Entered data:
Definition: steel: Creep Deformation sh: Deformation hardening steel su: Ultimate strain of steel Es: E-Module steel. fy: Yield stress steel. fu: Steel final effort
fy 4200kgf
cm2
⋅:= fu 6300kgf
cm2
⋅:= Es 29000 1000⋅lbf
in2
⋅:= Es 1.999 1011× Pa=
t
fu
fy
30 q⋅ 1+( )2⋅ 60 q⋅− 1−
15 q2⋅
:=
fs εs( ) fy−t εs− εsh−( )⋅ 2+
60 εs− εsh−( )⋅ 2+εs− εsh−( ) 60 t−( )⋅
2 30 q⋅ 1+( )2⋅
+
⋅ εs εsh<if
fy− εsh− εs≤ εsteel−<if
Es εs⋅ εsteel− εs≤ εsteel≤if
fy εsteel εs≤ εsh≤if
fyt εs εsh−( )⋅ 2+
60 εs εsh−( )⋅ 2+εs εsh−( ) 60 t−( )⋅
2 30 q⋅ 1+( )2⋅
+
⋅ εs εsh>if
:=
εsteelfy
Es:= εsteel 2.06 10
3−×=
εsu εsh 0.14+:=
εsh 16 εsteel⋅ εsh 0.033=
εsu 0.173=
q εsu εsh−:= q 0.14=
t 105.986=
Deformation in flow:
General formula for steel:
Deformation at the hardening:
Last Deformation steel:
Calculations for hardening steel:
0.1− 0 0.11− 10
9×
5− 108
×
0
5 108
×
1 109
×
fs εs( )
εs
γ εc( )2 250 εc⋅−( )
2 3 500 εc⋅−( )⋅εc .002≤if
1
56−
12 107⋅
6
10
εc 2⋅+100
3εc 3⋅−
εc26−
3 104⋅
12
10
εc⋅+ 50 εc 2⋅−
⋅− εc 0.002>if
:=
α εc( )10
3
6
3 εc⋅ 500 εc 2⋅−( )⋅ εc .002≤if
26−3 10000⋅ εc⋅
50 εc⋅−12
10+ εc 0.002>if
:=
Stress-strain graphic for steel – (Park and Paulay)
Geometric data
Entered data:
d11
d12
d21
d22
dn
h 200 cm⋅:= f́ c 280kgf
cm2
:=b 10 cm⋅:=
fy 4200kgf
cm2
:=ε c 0.003:=
ε su 0.002:=
d
194cm
194cm
170cm
150cm
130cm
110cm
90cm
70cm
50cm
30cm
6cm
6cm
:=
φ 1 4:=φ 2 4:=
φ 3 3:=φ 4 3:=φ 5 3:=φ 6 3:=φ 7 3:=φ 8 3:=φ 9 3:=φ 10 3:=φ 11 4:=φ 12 4:=
dmax max d( ) 194 cm⋅=:=
(Compressive strength)
(Yield strain steel)
(Strain concrete)
(Strain steel)
(Number diameter steel)
Calculation Steel-Areas:
Asi
π4
φ 1
8
2.54⋅ cm⋅
2
⋅
π4
φ 2
8
2.54⋅ cm⋅
2
⋅
π4
φ 3
8
2.54⋅ cm⋅
2
⋅
π4
φ 4
8
2.54⋅ cm⋅
2
⋅
π4
φ 5
8
2.54⋅ cm⋅
2
⋅
π4
φ 6
8
2.54⋅ cm⋅
2
⋅
π4
φ 7
8
2.54⋅ cm⋅
2
⋅
π4
φ 8
8
2.54⋅ cm⋅
2
⋅
π4
φ 9
8
2.54⋅ cm⋅
2
⋅
π4
φ 10
8
2.54⋅ cm⋅
2
⋅
π4
φ 11
8
2.54⋅ cm⋅
2
⋅
π4
φ 12
8
2.54⋅ cm⋅
2
⋅
:= Asi
1.267
1.267
0.713
0.713
0.713
0.713
0.713
0.713
0.713
0.713
1.267
1.267
cm2⋅=
Calculation: (Point 1) - Axial compression at zero moment
Balanced point: (Point 4)
Pmax 0.85 f́ c⋅ b h⋅
0
11
i
Asi i∑=
−
⋅
0
11
i
Asi i∑=
fy⋅+
:=
P max 518.661 tonnef=
εsi
cB d0−
cBεc⋅
cB d1−
cBεc⋅
cB d2−
cBεc⋅
cB d3−
cBεc⋅
cB d4−
cBεc⋅
cB d5−
cBεc⋅
cB d6−
cBεc⋅
cB d7−
cBεc⋅
cB d8−
cBεc⋅
cB d9−
cBεc⋅
cB d10−
cBεc⋅
cB d11−
cBεc⋅
:=
cB dmax
εc
εc εsu+⋅
116.4 cm⋅=:=
fsi
fs εsi0
fs εsi1
fs εsi2
fs εsi3
fs εsi4
fs εsi5
fs εsi6
fs εsi7
fs εsi8
fs εsi9
fs εsi10
fs εsi11
3-4.078·10
3-4.078·10
3-2.817·10
3-1.766·10
-714.667
336.314
31.387·10
32.438·10
33.489·10
34.2·10
34.2·10
34.2·10
kgf
cm2
⋅=:=εsi
-3-2·10
-3-2·10
-3-1.381·10
-4-8.66·10
-4-3.505·10
-41.649·10
-46.804·10
-31.196·10
-31.711·10
-32.227·10
-32.845·10
-32.845·10
=
Note: Getting more points by changing cB
Interaction Diagram Mathcad Type A: (Rectangular s hape)
0 50 100 150
200−
0
200
400
600
Pi
Mi
P0 α εc( ) fc⋅ b⋅ cB⋅
0
11
i
fsi iAsi i⋅
∑
=
+:=
P0 210.852 tonnef⋅=
M0 α εc( ) fc⋅ b⋅ cB⋅h
2γ εc( ) cB⋅−
⋅
0
11
i
fsi iAsi i⋅
h
2
di−
⋅
∑
=
+:=
M0 109.933 tonnef m⋅⋅=
P
518.661
253.023
210.852
168.682
105.426
63.255
42.17
12.09−
:= M
0
107.698
109.933
104.093
80.196
54.173
38.134
0
:=
Further Results with changed c: (Axial Force “P” and Bending Moment “M”)
BP
tonnef tonnef m
(Balanced condition)
(Balanced condition)
Comparison of the two calculations:
The area of “Type B” is twice as big as from “Type A”. But the wall with two columns can absorb about three to four times larger moments and Axial Forces. Partly because more steel is installed in the wall (Type B)
Results Shear Wall Type A: Rectangular shape (Excel)
Results Shear Wall Type B: Wall with two columns (Excel)
Pmax: Same value because of the same area in both cases
Points (Fragile-, Ductile-Condition and Balanced Point): Different values with Kent and Park and Paulay because of a more strict method of calculation including by using the functions of the E-Modules.
References:
Reinforced Concrete (A Fundamental Approach) - Edward G. Nawy 5th Edition
Reinforced concrete structures R.Park and T.Paulay
Seismic design aids for nonlinear analysis of reinforced concrete structures
ACI-318
NTE E.060
Results Shear Wall Type A: Rectangular shape (Excel)
Results Shear Wall Type A: Rectangular shape (Mathcad)
(Pmax)
(Pmax)